Suppose the a system has a Hamiltonian $H = H(q,p)$, and suppose $H$ does not depend explicitly on time. If $H\neq E$ the total energy of the system, does this necessarily say that $E$ is not conserved? Why?
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2This question has no general answer. Just from $H\neq E$ you cannot conclude that $E$ is not conserved. It depends on you particular system whether or not any expression that's not the Hamiltonian will be conserved. – ACuriousMind Jan 22 '16 at 02:04
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1Possible duplicate of Example where Hamiltonian $H \neq T+V=E$, but $E=T+V$ is conserved. Specifically, the answer at that question provides a counter example. – jacob1729 Nov 06 '19 at 19:57
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I think you have a logical contradiction in your question. When H is independent of time, then H=E.
Please refer to this previous answer. When is the Hamiltonian of a system not equal to its total energy?
Greg Petersen
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That is not correct. Consider the Lagrangian $L = \frac{1}{2}mR^2\left(\dot\theta^2 + \omega^2\sin^2\theta\right) - \frac{1}{2}k\left(L^2+R^2 - 2LR\cos\theta\right)$. It describes the motion of a mass on a ring, connected by a spring, and its Hamiltonian is independent in time, but is not the total energy. – JonTrav1 Jan 22 '16 at 08:11
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The post I flagged states, "In an ideal, holonomic and monogenic system (the usual one in classical mechanics), Hamiltonian equals total energy when and only when both the constraint and Lagrangian are time-independent and generalized potential is absent." Are you saying this hamiltonian violates this if and only if statement? If so we should revise this previous post. If not, what extra condition would need to be added to the H=E argument? Thanks for the feedback! – Greg Petersen Jan 22 '16 at 14:57